Question

(x-3)(x+5) больше или рано нулю.

Answer 1

(х-3)(х+5)=х^2-5х-3х-15 = х^2-8х+15 - больше нуля

Answer 2

To determine when the expression $(x-3)(x+5)$ is greater than or equal to zero, we need to find the intervals for which the expression is positive or zero.

To do this, we can use the concept of interval notation and test the expression for different intervals of $x$.

Step 1: Find the critical points (where the expression equals zero). To find the critical points, set $(x-3)(x+5)$ equal to zero and solve for $x$: $(x-3)(x+5) = 0$

Set each factor to zero and solve for $x$: $x - 3 = 0$ --> $x = 3$ $x + 5 = 0$ --> $x = -5$

Step 2: Test the intervals between the critical points.

Now, we have three intervals to consider: $(-\infty, -5)$, $(-5, 3)$, and $(3, \infty)$.

Step 3: Pick a test point from each interval and evaluate the expression.

For $(x-3)(x+5) > 0$:

• Pick a test point $x_1$ from $(-\infty, -5)$, for example, $x_1 = -6$. $(x_1 - 3)(x_1 + 5) = (-6 - 3)(-6 + 5) = (-9)(-1) = 9 > 0$

• Pick a test point $x_2$ from $(-5, 3)$, for example, $x_2 = 0$. $(x_2 - 3)(x_2 + 5) = (0 - 3)(0 + 5) = (-3)(5) = -15 < 0$

• Pick a test point $x_3$ from $(3, \infty)$, for example, $x_3 = 4$. $(x_3 - 3)(x_3 + 5) = (4 - 3)(4 + 5) = (1)(9) = 9 > 0$

Step 4: Determine the intervals where $(x-3)(x+5) > 0$.

The expression $(x-3)(x+5)$ is greater than zero ($> 0$) in the intervals $(-5, 3)$ and $(3, \infty)$.

Step 5: Include the endpoints where $(x-3)(x+5) \geq 0$.

Since the original question asks for "greater than or equal to zero," we need to include the critical points as well.

The expression $(x-3)(x+5)$ is greater than or equal to zero ($\geq 0$) in the intervals $[-5, 3]$ and $(3, \infty)$.

In interval notation, the solution is: $[-5, 3] \cup (3, \infty)$